ECE4762011_Lect21

# ECE4762011_Lect21 - Lecture 21 Symmetrical Components...

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Unformatted text preview: Lecture 21 Symmetrical Components, Unbalanced Fault Analysis Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS 2 Announcements Be reading Chapters 9 and 10 HW 8 is due now. HW 9 is 8.4, 8.12, 9.1,9.2 (bus 2), 9.14; due Nov 10 in class Start working on Design Project. Tentatively due Nov 17 in class Second exam is on Nov 15 in class. Same format as first exam, except you can bring two note sheets (e.g., the one from the first exam and another) 3 Single Line-to-Ground (SLG) Faults Unbalanced faults unbalance the network, but only at the fault location. This causes a coupling of the sequence networks. How the sequence networks are coupled depends upon the fault type. We’ll derive these relationships for several common faults. With a SLG fault only one phase has non-zero fault current -- we’ll assume it is phase A. 4 SLG Faults, cont’d ? f a f b f c I I I = 2 2 Then since 1 1 1 ? 1 1 1 3 3 1 f f f f f f a f I I I I I I I α α α α + +-- = → = = = 5 SLG Faults, cont’d 2 2 1 1 1 1 1 This means The only way these two constraints can be satisified is by coupling the sequence networks in series f f a f a f f a f f b f c f f a f f f V Z I V V V V V V V V V V α α α α +- +- = = = + + 6 SLG Faults, cont’d With the sequence networks in series we can solve for the fault currents (assume Z f =0) 1.05 0 1.964 (0.1389 0.1456 0.25 3 ) 5.8 (of course, 0) f f f f f f f s a c b I j I I j Z I j I I +- ∠ ° = = - = = + + + = → = - = = I A I 7 Example 9.3 8 Line-to-Line (LL) Faults The second most common fault is line-to-line, which occurs when two of the conductors come in contact with each other. With out loss of generality we'll assume phases b and c. bg Current Relationships: 0, , Voltage Relationships: V f f f a c f b cg I I I I V = = - = = 9 LL Faults, cont'd ( 29 ( 29 2 2 2 2 Using the current relationships we get 1 1 1 1 1 3 1 1 1 3 3 Hence f f f b f b f f f f f f b b f f I I I I I I I I I I I I α α α α α α α α +- +- +- = → - = =- =- = - 10 LL Faults, con'td ( 29 ( 29 2 2 2 2 Using the voltage relationships we get 1 1 1 1 1 3 1 Hence 1 3 1 3 f f ag f f bg f f cg f f f ag bg f f f ag f f bg V V V V V V V V V V V V V V α α α α α α α α +- +- +-...
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ECE4762011_Lect21 - Lecture 21 Symmetrical Components...

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